In \(\mathrm{CF}_{3} \mathrm{Cl}\) the \(\mathrm{C}-\mathrm{Cl}\) bond- dissociation energy is 339 \(\mathrm{kJ} / \mathrm{mol} .\) In \(\mathrm{CCl}_{4}\) the \(\mathrm{C}-\mathrm{Cl}\) bond dissociation energy is 293 \(\mathrm{kJ} / \mathrm{mol} .\) What is the range of wavelengths of photons that can cause \(\mathrm{C}-\mathrm{Cl}\) bond rupture in one molecule but not in the other?

First, we will find the difference in bond dissociation energies between CF3Cl and CCl4: ΔE = |Bond dissociation energy of CF3Cl - Bond dissociation energy of CCl4| ΔE = |339 kJ/mol - 293 kJ/mol| ΔE = 46 kJ/mol

Next, we will convert the energy difference from kJ/mol to J/mol, as the constants in the Planck's equation are in Joules. 1 kJ = 1000 J, so the conversion is: ΔE = 46 kJ/mol * 1000 J/1 kJ = 46000 J/mol

We will use the Planck's equation to find the range of wavelengths that corresponds to the energy difference calculated in step 2: E = hc/λ Here, E is the energy difference, h is the Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength. Let us first determine the maximum wavelength (λ_max) that can cause a bond rupture in CF3Cl: E = 339 kJ/mol * 1000 J/1 kJ = 339000 J/mol λ_max = hc/E = (6.626 x 10^-34 Js)(3.0 x 10^8 m/s) / (339000 J/mol) Now, we will determine the minimum wavelength (λ_min) that can cause a bond rupture in CCl4: E = 293 kJ/mol * 1000 J/1 kJ = 293000 J/mol λ_min = hc/E = (6.626 x 10^-34 Js)(3.0 x 10^8 m/s) / (293000 J/mol) Finally, calculating λ_max and λ_min: λ_max = 5.874 x 10^-19 m/mol λ_min = 6.788 x 10^-19 m/mol

To express the range of wavelengths in nanometers (nm), we will convert the values in meters to nanometers (1 m = 10^9 nm): λ_max = (5.874 x 10^-19 m/mol) * (10^9 nm/1 m) = 587.4 nm λ_min = (6.788 x 10^-19 m/mol) * (10^9 nm/1 m) = 678.8 nm

The range of wavelengths of photons that can cause C-Cl bond rupture in one molecule but not in the other is: λ_min < λ < λ_max 678.8 nm < λ < 587.4 nm